Let's write the balanced nuclear equations step-by-step for each of the given processes.
### (a) Alpha Emission from Curium-242
An alpha particle ([tex]\(_2^4He\)[/tex]) is emitted from curium-242 ([tex]\(_{96}^{242}Cm\)[/tex]). In alpha emission, the nucleus loses 2 protons and 2 neutrons. Therefore, the resulting element will have 2 fewer protons (which means it will shift 2 places back in the periodic table) and its mass number will decrease by 4.
The equation is:
[tex]\[ \mathrm{_{96}^{242}Cm} \rightarrow \mathrm{_{2}^{4}He} + \mathrm{_{Z}^{A}X} \][/tex]
Where:
- [tex]\(Z = 96 - 2 = 94\)[/tex]
- [tex]\(A = 242 - 4 = 238\)[/tex]
We identify the resulting element as plutonium ([tex]\(_{94}^{238}Pu\)[/tex]).
So, the balanced nuclear equation is:
[tex]\[ \mathrm{_{96}^{242}Cm} \rightarrow \mathrm{_{2}^{4}He} + \mathrm{_{94}^{238}Pu} \][/tex]
### (b) Beta Emission from Magnesium-28
In beta emission, a neutron in the nucleus is converted into a proton, an electron (beta particle), and an antineutrino. The number of protons increases by one, while the mass number remains the same.
The equation is:
[tex]\[ \mathrm{_{12}^{28}Mg} \rightarrow \mathrm{_{-1}^{0}e} + \mathrm{_{Z}^{A}X} \][/tex]
Where:
- [tex]\(Z = 12 + 1 = 13\)[/tex]
- [tex]\(A = 28\)[/tex]
We identify the resulting element as aluminum ([tex]\(_{13}^{28}Al\)[/tex]).
So, the balanced nuclear equation is:
[tex]\[ \mathrm{_{12}^{28}Mg} \rightarrow \mathrm{_{-1}^{0}e} + \mathrm{_{13}^{28}Al} \][/tex]
### (c) Positron Emission from Xenon-118
In positron emission, a proton in the nucleus is converted into a neutron, a positron (positive electron), and a neutrino. The number of protons decreases by one, while the mass number remains the same.
The equation is:
[tex]\[ \mathrm{_{54}^{118}Xe} \rightarrow \mathrm{_{1}^{0}e} + \mathrm{_{Z}^{A}X} \][/tex]
Where:
- [tex]\(Z = 54 - 1 = 53\)[/tex]
- [tex]\(A = 118\)[/tex]
We identify the resulting element as iodine ([tex]\(_{53}^{118}I\)[/tex]).
So, the balanced nuclear equation is:
[tex]\[ \mathrm{_{54}^{118}Xe} \rightarrow \mathrm{_{1}^{0}e} + \mathrm{_{53}^{118}I} \][/tex]
